\(n_{CuO}=\dfrac{32}{80}=0,4(mol)\\ CuO+2HCl\to CuCl_2+H_2\\ \Rightarrow n_{HCl}=0,8(mol);n_{CuCl_2}=n_{H_2}=0,4(mol)\\ a,m_{dd_{HCl}}=\dfrac{0,8.36,5}{20\%}=146(g)\\ b,m_{CuCl_2}=0,4.135=54(g)\\ c,C\%_{CuCl_2}=\dfrac{54}{32+146-0,4.2}.100\%=30,47\%\)

\(CuO + 2HCl \rightarrow CuCl_2 + H_2O\)

\(n_{CuO}= \dfrac{32}{80}= 0,4 mol\)

Theo PTHH:

\(n_{HCl}= 2n_{CuO}= 0,8 mol\)

\(\Rightarrow m_{HCl}= 0,8 . 36,5=29,2 g\)

\(\rightarrow m_{dd HCl}= \dfrac{29,2 . 100%}{20%}= 146 g\)

b) Muối tạo thành là CuCl₂

Theo PTHH:

\(n_{CuCl_2}= n_{CuO}= 0,4 mol\)

\(\Rightarrow m_{CuCl_2}= 0,4 . 135= 54g\)

c)

\(m_{dd sau pư}= m_{CuO} + m_{dd HCl}= 32 + 146=178 g\)

C%= \(\dfrac{54}{178} . 100\)%= 30,337 %

\(-\) \(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)

\(pt:CuO+2HCl\rightarrow CUCl_2+H_2O\)

 \(a,\) 

\(-\) \(n_{HCl}=2.n_{CuO}=2.0,4=0,8\left(mol\right)\)

\(\rightarrow\) \(m_{HCl}=0,8.36,5=29,2\left(gam\right)\)

\(\Rightarrow\) \(m_{ddHCl}=29,2:20\%=146\left(gam\right)\)

\(b,\)

\(-\) \(n_{CuCl_2}=n_{CuO}=0,4\left(mol\right)\)

\(\rightarrow\) \(m_{CuCl_2}=0,4.135=54\left(gam\right)\)

\(c,\)

\(m_{ddsaupu}=32+146=178\left(gam\right)\)

\(\Rightarrow\) \(C\%_{CuCl_2}=\dfrac{54}{178}.100\%=30,34\%\)

Zn + 2HCl --> ZnCl2 + H2

0,5      1              0,5       0,5

nZn=\(\dfrac{32,5}{65}\)=0,5(mol)

mZnCl2=0,5.136=68(g)

VH2=0,5.22,4=11,2(l)

c) CM HCl=\(\dfrac{1}{0,25}=4M\)

Bài 1

n_BaCl2= 200 *2.6%= 5.2 (g) ; n_BaCl2= 5.2/208=0.025(mol)

n_H2SO4=49*10%=4.9(g) ; n_H2SO4=4.9/98=0.05(mol)

PTHH

..........................H₂SO₄ + BaCl₂ ➞ 2HCl + BaSO₄

Trước phản ứng:0.05 : 0.025...................................(mol)

Trong phản ứng:0.025 : 0.025......... : 0.025 : 0.05(mol)

Sau phản ứng : 0.025 : 0 ......... : 0.025 : 0.05 (mol)

a) m_BaSO4=0.025*233=5.825(g)

b) m_{dd sau phản ứng} = 49+200-5.825=243.175(g)

C% (H₂SO₄) = (0.025* 98)/243.175*100%=1.007%

C% (HCl) = (0.05*36.5)/243.175*100%=0.007%

Bài 2:

n_HCl= 73 *25%= 18.25 (g) ; n_HCl= 18.25/36.5=0.5(mol)

n_AgNO3=34*5%=1.7(g) ; n_AgNO3=1.7/170=0.01(mol)

PTHH

..........................HCl + AgNO₃ ➞ AgCl + 2HNO₃

Trước phản ứng:0.5 : 0.01......................................(mol)

Trong phản ứng:0.01 : 0.01.............. : 0.01 : 0.01(mol)

Sau phản ứng : 0.49: 0 ............... : 0.01 : 0.01(mol)

a) m_AgCl=0.01*143.5=1.435(g)

b) m_{dd sau phản ứng} = 73+34-1.435=105.565(g)

C% (HNO₃) = (0.01* 63)/105.565*100%=0.0059%

C% (HCl) = (0.49*36.5)/105.565*100%=16.94%

Bài 1: \(n_{H_2SO_4}=\frac{9}{49}\left(mol\right)\)

H₂SO₄ + 2KOH -> K₂SO₄ + 2H₂O

=> n_KOH= 2n_H2SO4 = \(\frac{18}{49}\left(mol\right)\)

=> V_{dd KOH} = \(\frac{18}{49}:\frac{2}{1000}=\frac{9000}{49}\left(ml\right)\)

b) n_K2SO4 = n_H2SO4 = \(\frac{9}{49}\left(mol\right)\)

=> m_K2SO4= \(\frac{9}{49}\cdot174=\frac{1566}{49}\left(g\right)\)

m_{dd KOH} = \(\frac{9000}{49}\cdot1,12=\frac{1440}{7}\left(g\right)\)

c) \(\%m_{K_2SO_4}=\frac{1566}{49}:\left(200+\frac{1440}{7}\right)\cdot100\%\approx7,87\%\)

bài 2: n_Na2CO3 = 0,05 (mol)

PTHH:

Na₂CO₃ + 2HCl -> 2NaCl + H₂O + CO₂

=> n_HCl = n _NaCl = 2n_Na2CO3 = 0,1 (mol)

=> m_NaCl= 0,1 . 58,5 = 5,85 (g)

b) n_CO2 = n_Na2CO3 = 0,05 (mol)

=> m_CO2 = 0,05 . 44 = 2,2 (g)

m_{dd HCl} = 0,1 . 36,5 :20% = 18,25 (g)

=> %m_NaCl = \(\frac{5,85}{53+18,25-2,2}\approx8,47\%\)

Bài 3; n_AgNO3 = 0,3 (mol)

n_HCl = 0,25 (mol)

PTHH:

AgNO₃ + HCl -> AgCl + HNO₃

Tỷ lệ: \(\frac{0,3}{1}>\frac{0,25}{1}\)=> AgNO₃ dư tính theo HCl

=> n_{AgNO3 (p/ứ)} = n_AgCl = n_HNO3 = 0,25 (mol)

=> m_AgCl = 0,25 . 143,5 = 35,875 (g)

b) n_{AgNO3 dư} = 0,3- 0,25 = 0,05 (mol)

Cm_{AgNO3 dư} = \(\frac{0,05}{300+500}\cdot1000=0,0625M\)

Cm_HNO3 = \(\frac{0,25}{300+500}\cdot1000=0,3125M\)

\(n_{NaOH}=\dfrac{200.5\%}{100\%.40}=0,25(mol)\\ n_{HCl}=\dfrac{36,5.20\%}{100\%.36,5}=0,2(mol)\\ a,NaOH+HCl\to NaCl+H_2O\)

Vì \(\dfrac{n_{NaOH}}{1}>\dfrac{n_{HCl}}{1}\) nên \(NaOH\) dư

\(b,n_{NaOH(dư)}=0,25-0,2=0,05(mol);n_{NaCl}=0,2(mol)\\ \Rightarrow m_{\text{dd sau p/ứ}}=0,2.58,5+0,05.40=13,7(g)\\ c,m_{NaCl}=0,2.58,5=11,7(g)\\ d,n_{H_2}=0,2(mol)\\ \Rightarrow \begin{cases} C\%_{NaOH}=\dfrac{0,05.40}{36,5+200-0,2.2}.100\%=0,85\%\\ C\%_{NaCl}=\dfrac{11,7}{36,5+200-0,2.2}.100\%=4,96\% \end{cases}\)

Bài 1 :

PTHH : Zn + H₂SO₄ ------> ZnSO₄ + H₂

\(n_{Zn}=\frac{m}{M}=\frac{26}{65}=0,4\left(mol\right)\)

Theo PTHH : n_H2 = n_Zn = 0,4 mol

=> Khối lượng H₂ được tạo ra ở đktc là :

\(V=n\times22,4\)

\(\Rightarrow V_{H_2}=0,4\times22,4\)

\(\Rightarrow V_{H_2}=8,96\left(l\right)\)

Theo PTHH : n_ZnSO4 = n_Zn = 0,4 mol

=> Khối lượng muối được tạo thành là :

\(m=n\times M\)

\(\Rightarrow m_{ZnSO_4}=0,4\times161\)

\(\Rightarrow m_{ZnSO_4}=64,4\left(g\right)\)

b) Theo PT : \(n_{H_2SO_4}=n_{Zn}=0,4\left(mol\right)\)

=> Khối lượng \(H_2SO_4\)cần dùng cho phản ứng là :

\(m=n\times M\)

\(\Rightarrow m_{H_2SO_4}=0,4\times98\)

\(\Rightarrow m_{H_2SO_4}=39,2\left(g\right)\)

c) Nồng độ phần trăm thu được sau phản ứng là :

\(C\%=\frac{m_{ct}}{m_{dd}}\times100\%\)

\(\Rightarrow C\%=\frac{39,2}{64,4}\times100\%\approx60,9\%\)

Vậy :.........................

mKOH=28(g)

nKOH=0.5(mol)

PTHH:2KOH+H2SO4->K2SO4+2H2O

a)Theo pthh:nH2SO4=1/2 nKOH->nH2SO4=0.25(mol)

mH2SO4=0.25*98=24.5(g)

C%ddH2SO4=24.5/100*100=24.5%

theo pthh:nK2SO4=nH2SO4->nK2SO4=0.25(mol)

mK2SO4=0.25*(39*2+96)=43.5(g)

c)mdd sau phản ứng:200+100=300(g)

d) C% muối=43.5:300*100=14.5%